document.write( "Question 1198404: In hw many ways can three items be selected from a group of six items. Use letters A B C D E and F to identity the items and list each of the different combination of three items \n" ); document.write( "
Algebra.Com's Answer #831996 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "There are 6 choices for the first slot, then 6-1 = 5 choices for the second slot, and finally 5-1 = 4 choices for the third slot.\r
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\n" ); document.write( "\n" ); document.write( "If order mattered, then we'd have 6*5*4 = 30*4 = 120 different permutations, and this would be the final answer.\r
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\n" ); document.write( "\n" ); document.write( "But order doesn't matter.
\n" ); document.write( "The group {A,B,C} is the same as something like {B,A,C}.
\n" ); document.write( "The order doesn't matter when considering the group overall.\r
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\n" ); document.write( "\n" ); document.write( "There are 3*2*1 = 6 ways to rearrange any given group of 3 items.
\n" ); document.write( "This tells us that the previous figure of 120 is too large by a factor of 6.\r
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\n" ); document.write( "\n" ); document.write( "Divide by 6 to correct for this over-counting: 120/6 = 20
\n" ); document.write( "Therefore, we have 20 different groups of three letters chosen from a pool of six.\r
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\n" ); document.write( "\n" ); document.write( "Alternatively you can use the nCr combination formula to get this answer.
\n" ); document.write( "\"nCr+=+%28n%21%29%2F%28r%21%2A%28n-r%29%21%29\"
\n" ); document.write( "Use n = 6 and r = 3. \r
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\n" ); document.write( "\n" ); document.write( "Another alternative is to use Pascal's Triangle.
\n" ); document.write( "Look at the row that starts with 1,6,... and you're looking for the item in the fourth slot.
\n" ); document.write( "This is because the 'r' value starts counting at r = 0 rather than r = 1.\r
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\n" ); document.write( "\n" ); document.write( "Answer: 20
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